Chapter 1 – Algebra I
Solutions to Exercise 1A
1 a Add indices: x3⇥ x4⇥x3+4 = x7 b Add indices:
a5⇥a 3 =a5+ 3 =a2 c Add indices:
x2⇥ x 1⇥x2 = x2+ 1+2 = x3 d Subtract indices:
y3
y7 =y3 7 =y 4 e Subtract indices:
x8
x 4 = x8 ( 4) = x12 f Subtract indices:
p 5
p2 = p 5 2= p 7 g Subtract indices:
a12 ÷a23 =a36 46 =a 16 h Multiply indices:
(a 2)4 = a 2⇥4 = a 8 i Multiply indices:
(y 2) 7 = y 2⇥( 7)= y14 j Multiply indices:
(x5)3 = x5⇥3 = x15 k Multiply indices:
(a 20)35 = a 20⇥35 =a 12 l Multiply indices:
✓ x 12◆
4
= x 12⇥ 4 = x2 m Multiply indices:
(n10)15 = n10⇥15 =n2
n Multiply the coefficients and add the indices:
2x12 ⇥4x3 = (2⇥4)x12+3 =8x72 o Multiply the first two indices and add
the third:
(a2)52 ⇥a 4 = a2⇥52 ⇥a 4
= a5+( 4) = a1 =a
p 1
x 4 = x1÷ 1 4 = x4
q ✓
2n 25◆5÷(43n4)=25n 25⇥5÷((22)3n4)
=25n 2÷(26n4) =25 6n 2 4 =2 1n 6 = 1
2n6 r Multiply the coefficients and add the
indices.
x3⇥2x12 ⇥ 4x 32
= (1⇥2⇥ 4)x3+
1 2+
✓ 3 2 ◆
= 8x2
s (ab3)2⇥a 2b 4⇥ 1 a2b 3
= a2b6⇥a 2b 4⇥a 2b3 = a2+ 2+ 2b6+ 4+3 = a 2b5
t (22p 3 ⇥43p5÷((6p 3))0 = 1 Anything to the power zero is 1.
uncorrected
2 a 2512 = p25=5 b 6413 = p364= 4
c ✓16 9
◆1 2
= 16
1 2 912
=
p 16 p
9 = 4 3 d 16 12 = 1
1612
= p1
16 = 1 4
e ✓49 36
◆ 1 2
= 1
✓49 36
◆1 2
= p1
49 p
36
=
p 36 p
49 = 6 7 f 2713 = p327= 3
g 14412 = p144=12
h 6423 =✓6413◆2 = 42 =16 i 932 =✓912◆3
=33 = 27
j ✓81 16
◆14
= 81
1 4 1614
= 3
2
k ✓23 5
◆0
=1
l 12837 =
✓ 12817◆3
=23 = 8
3 a 4.352 =18.9225⇡18.92 b 2.45 =79.62624⇡79.63
c p34.6921= 5.89
d 0.02 3 =125 000 e p3
0.729=0.9
f p4
2.3045=1.23209. . .⇡ 1.23 g (345.64) 13 = 0.14249. . . ⇡0.14 h (4.558)25 = 1.83607. . . ⇡1.84
i 1
(0.064) 13
=(0.064)13 = 0.4 4 a aa22bb34 =a2 2b3 4
=a4b7
b 2a2(2b)3 (2a) 2b 4 =
2a2⇥23b3 2 2a 2b 4
= 2
4a2b3 2 2a 2b 4
=24 2a2 2b3 4 =26a4b7 = 64a4b7
c aa 22bb 34 =a 2 2b 3 4 =a0b1 = b
uncorrected
d aa22bb34 ⇥ a ab1b 1
= a
2+1b3+1 a 2+ 1b 4+ 1
= a
3b4 a 3b 5
=a3 3b4 5 = a6b9
e (2a)2⇥8b3 16a 2b 4 =
4a2⇥8b3 16a 2b 4
= 32a
2b3 16a 2b 4
= 32
16a2 2b3 4
= 2a4b7
f 2a2b3 8a 2b 4 ÷
16ab (2a) 1b 1
= 2a
2b3 8a 2b 4 ⇥
(2a) 1b 1 16ab
= 2a
2b3 8a 2b 4 ⇥
2 1a 1b 1 16ab
= 2
1+ 1a2+ 1b3+ 1 8⇥16⇥a 2+1b 4+1
= 2
0a1b2 128a 1b 3
= 1
128a1 1b2 3 = a2b5
128
5 2n⇥8n 22n ⇥16 =
2n⇥(23)n
22n⇥24
= 2 n⇥23n
22n⇥24
= 2 n+3n 2n
24
=22n ⇥2 4 =22n 4
6 2 x⇥3 x⇥62x⇥32x⇥22x
= (2⇥3) x⇥62x⇥(2⇥3)2x = 6 x⇥62x⇥62x
= 6 x+2x+2x = 63x
7 In each case, add the fractional indices. a 213 ⇥216 ⇥2 32 =226+16+ 46
=2 16 =✓1 2
◆16
b a14 ⇥a25 ⇥a 101 = a205+208+ 202
= a1120 c 223 ⇥256 ⇥2 32 =246+56+ 46
=256 d ✓213◆2⇥✓221◆5 = 223 ⇥252
= 246+156 =2196 e ✓213◆2⇥213 ⇥2 52 = 223 ⇥213 ⇥2 25
= 223+13+ 25 = 235 8 a
3
p
a3b2÷ p3a2b 1 =(a3b2)13 ÷(a2b 1)13
=a1b23 ÷a23b 13
=a1 23b23 13 =a13b
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b pa3b2⇥ pa2b 1
= (a3b2)12 ⇥(a2b 1)12
= a32b1⇥a1b 12
= a32+1b1+ 12 =a52b12 c
5
p
a3b2⇥ p5a2b 1 = (a3b2)15 ⇥(a2b 1)15
= a35b25 ⇥a25b 15
= a35+25b25+ 15 =ab15 d pa 4b2⇥ pa3b 1
= (a 4b2)12 ⇥(a3b 1)12
= a 2b1⇥a32b 12
= a 2+32b1+ 12
= a 12b12
= b
1 2 a12
=
✓b a
◆1 2
e pa3b2c 3⇥ pa2b 1c 5
= (a3b2c 3)12 ⇥(a2b 1c 5)12
= a32b1c 32 ⇥a1b 12 c 52
= a32+1b1+ 12c 32+ 52
= a52b12c 4 f p5a3b2÷ p5a2b 1
= (a3b2)15 ÷(a2b 1)15
= a35b25 ÷a25b 15
= a35 25b25 15 =a15b35
g p
a3b2 a2b 1c 5 ⇥
p a 4b2 a3b 1 ⇥
p a3b 1
= (a
3b2)12 a2b 1c 5 ⇥
(a 4b2)12
a3b 1 ⇥(a3b 1) 1 2
= a
3 2b1 a2b 1c 5 ⇥
a 2b1 a3b 1 ⇥a
3 2b 12
= a32 2b1 1c0 5⇥a 2 3b1 1 ⇥a32b 12
= a 12b2c5⇥a 5b2⇥a32b 12
= a 12+ 5+32b2+2+ 12c5
= a 4b72c5
uncorrected
Solutions to Exercise 1B
1 a 47.8=4.78⇥101 = 4.78⇥10b 6728= 6.728⇥103
c 79.23=7.923⇥101 = 7.923⇥10 d 43 580= 4.358⇥104
e 0.0023=2.3⇥10 3 f 0.000 000 56=5.6⇥10 7 g 12.000 34= 1.2000 34⇥101
= 1.2000 34⇥10
h Fifty million = 50 000 000 = 5.0⇥107 i 23 000 000 000= 2.3⇥1010
j 0.000 000 0013=1.3⇥10 9
k 165 thousand = 165 000 = 1.65⇥105
l 0.000 014 567=1.4567⇥10 5
2 a The decimal point moves 8 places to the right =1.0⇥10 8
b The decimal point moves 24 places to the right = 1.67⇥10 23
c The decimal point moves 5 places to the right = 5.0⇥10 5
d The decimal point moves 3 places to the left = 1.853 18⇥103
e The decimal point moves 12 places to the left = 9.461⇥1012
f The decimal point moves 10 places to the right = 2.998⇥1010
3 a The decimal point move 13 places to the right = 81 280 000 000 000
b The decimal point move 8 places to the right =270 000 000
c The decimal point move 13 places to the left =0.000 000 000 000 28 4 a 456.89⇡4.569⇥102
(4 significant figures) b 34567.23⇡3.5⇥104 (2 significant figures) c 5679.087⇡5.6791⇥103
(5 significant figures) d 0.04536⇡4.5⇥10 2 (2 significant figures) e 0.09045⇡9.0⇥10 2 (2 significant figures) f 4568.234⇡4.5682⇥103
(5 significant figures)
5 a 324 000⇥0.000 000 7 4000
= 3.24⇥10
5⇥7⇥10 7 4⇥103
= 3.24⇥7
4 ⇥105+ 7 3
= 5.67⇥10 5 = 0.0000567
uncorrected
b 5 240 000⇥0.8 42 000 000
= 5.24⇥10
6⇥8⇥10 1 4.2⇥107
= 41.92 ⇥10
5 4.2⇥107
= 4192 ⇥10
3 42 000⇥103
= 4192
42 000 = 262 2625
6 a pb34a =
3
p
2⇥109 3.2154
=
3
p
2⇥ p3109 106.8375. . .
= 1.2599. . .⇥10
3 106.8375. . .
=0.011 792 . . . ⇥103 ⇡ 11.8 b
4
pa 4b4 =
4
p
2⇥1012 4⇥0.054
=
4
p
2⇥ p41012 4⇥0.000 006 25
= 1.189 2. . . ⇥10
3 4⇥6.25⇥10 6
=0.047 568. . .⇥109 ⇡ 4.76⇥107
uncorrected
Solutions to Exercise 1C
1 a 3x+7=153x=15 7 =8
x= 8
3
b 8 x
2 = 16 x
2 = 16 8
= 24
x
2 ⇥ 2= 24⇥ 2 x= 48
c 42+3x=22
3x=22 42 = 20
x= 20
3 d 2x
3 15= 27 2x
3 = 27+15
= 42
2x 3 ⇥
3
2 = 42⇥ 3 2 x= 63
e 5(2x+4)=13
10x+20=13
10x=13 20 = 7
x= 7
10 = 0.7
f 3(4 5x)=24
12+15x=24
15x=24+12 =36
x= 36
15
= 12
5 =2.4 g 3x+5= 8 7x
3x+7x= 8 5
10x= 3
x= 3
10 = 0.3 h 2+3(x 4)=4(2x+5)
2+3x 12=8x+20
3x 10=8x+20
3x 8x=20+10
5x=30
x= 30
5 = 6
i 2x
5 3 4 =5x 2x
5 ⇥20 3
4 ⇥20=5x⇥20 8x 15=100x
8x 100x =15
92x =15
x = 15
92
uncorrected
j 6x+4= x
3 3
6x⇥3+4⇥3= x
3⇥3 3⇥3 18x+12= x 9
18x x= 9 12
17x= 21
x= 21
17
2 a x
2 + 2x
5 =16 x
2 ⇥10+ 2x
5 ⇥10=16⇥10 5x+4x=160
9x=160
x= 160
9
b 3x
4 x 3 =8 3x
4 ⇥12 x
3 ⇥12=8⇥12 9x 4x=96
5x=96
x= 96
5 =19.2
c 3x 2
2 +
x 4 = 8 3x 2
2 ⇥4+ x
4 ⇥4= 8⇥4 2(3x 2)+x= 32
6x 4+x= 32
7x= 32+4 = 28
x= 4
d 5x
4 4 3 =
2x 5 5x
4 ⇥60 4
3 ⇥60= 2x
5 ⇥60 75x 80=24x
75x 24x =80
51x =80
x = 80
51
e x 4
2 +
2x+5
4 = 6
x 4 2 ⇥4+
2x+5
4 ⇥4= 6⇥4 2(x 4)+(2x+5)= 24
2x 8+2x+5= 24
4x= 24+8 5 = 27
x= 27
4 = 6.75 f
3 3x 10
2(x+5)
6 =
1 20 3 3x
10 ⇥60
2(x+5)
6 ⇥60= 1 20⇥60 6(3 3x) 20(x+5)= 3
18 18x 20x 100= 3
38x= 3 18+100 = 85
x= 85
38
uncorrected
g 34x 2(x+1)
5 = 24
3 x
4 ⇥20
2(x+1)
5 ⇥20= 24⇥20
5(3 x) 8(x+1)= 480
15 5x 8x 8= 480
13x= 480 15+8
= 487
x=487
13
h 2(58 x)+6
7= 4(x 2)
3 2(5 x)
8 ⇥168+
6 7⇥168=
4(x 2)
3 ⇥168
42(5 x)+144=224(x 2)
210+42x+144=224x 448
42x 224x= 448+210 144
182x= 382
x=382
182 =
191 91
3 a 3x+2y=2; 2x 3y=6
Use elimination. Multiply the first equation by 3 and the second equation by 2.
9x+6y=6 1
4x 6y=12 2
1 + 2: 13x=18
x= 18
13
Substitute into the first equation: 3⇥ 1813 +2y=2
54
13 +2y=2 2y=2 54
13
= 28
13 y= 14
13
b 5x+2y= 4; 3x y =6
Use elimination. Multiply the second equation by 2.
5x+2y=4 1
6x 2y=12 2
1 + 2: 11x=16
x= 16
11
Substitute into the second, simpler equation:
3⇥ 1611 y=6
48
11 y=6 y=6 48
11 y= 18
11 c 2x y= 7; 3x 2y=2
Use substitution. Makeythe subject of the first equation.
y= 2x 7
Substitute into the second equation: 3x 2(2x 7)=2
3x 4x+14=2
x=2 14
x=12
Substitute into the equation in which yis the subject:
y= 2⇥12 7 = 17
d x+2y= 12;x 3y=2
Use substitution. Make xthe subject of the first equation.
x=12 2y
uncorrected
Substitute into the second equation: 12 2y 3y=2
5y=2 12 = 10
y=2
Substitute into the first equation: x+2⇥2=12
x+4=12
x=8
e 7x 3y= 6;x+5y=10
Use substitution. Make xthe subject of the second equation.
x= 10 5y
Substitute into the first equation: 7(10 5y) 3y= 6
70 35y 3y= 6
38y= 6 70 = 76
y= 76
38 =2
Substitute into the second equation: x+5⇥2=10
x+10=10
x=0
f 15x+2y= 27; 3x+7y=45
Use elimination. Multiply the second equation by 5.
15x+2y= 27 1
15x+35y= 225 2
1 2:
33y= 198
y= 198
33 =6
Substitute into the second equation: 3x+7⇥6=45
3x+42=45
3x=45 42 =3
x=1
uncorrected
Solutions to Exercise 1D
1 a 4(x 2)=604x 8=60
4x=60+8 =68
x=17
b The length of the square is 2x+7 4 . 2x+7
4 !2
=49
2x+7
4 =7
2x+7=7⇥4=28
2x=28 7 =21
x=10.5
c The equation is length=twice width.
x 5=2(12 x)
x 5=24 2x
x+2x =24+5
3x =29
x = 29
3
d y= 2((2x+1)+(x 3)) = 2(2x+1+x 3) = 2(3x 2)
= 6x 4
e Q=np
f If a 10% service charge is added, the total price will be multiplied by 110%, or 1.1.
R= 1.1pS
g Using the fact that there are 12 lots of 5 min in an hour (60÷12=5),
60n
5 =2400
h a= circumference ⇥ 60
360
= 2⇡(x+3)⇥ 60
360
= 2⇡(x+3)⇥ 1
6
= ⇡
3(x+3)
2 Let the value of Bronwyn’s sales in the first week be $s.
s+( s +500)+( s +1000) +( s +1500)+( s +2000) =17500
5s+5000=17500
5s=12500
s=2500
The value of her first week’s sales is $2500.
3 Letdbe the number of dresses bought andhthe number of handbags bought.
65d+26h= 598
d+h= 11
Multiply the second equation by 26 (the smaller number).
65d+26h= 598 1 26d+26h= 286 2
uncorrected
1 2: 39d =312
d = 312
39 =8 h+8=11
h=3
Eight dresses and three handbags.
4 Let the courtyard’s width bewmetres. 3w+w+3w+w=67
8w=67
w=8.375
The width is 8.375 m.
The length= 3⇥8.375=25.125 m.
5 Let pbe the full price of a case of wine. The merchant will pay 60% (0.6) on the 25 discounted cases.
25p+25⇥0.6p= 2260
25p+15p= 2260
40p= 2260
p= 56.5
The full price of a case is $56.50.
6 Let xbe the number of houses with an $11 500 commission andybe the number of houses with a $13 000 commission.
We only need to findx. x+y=22
11 500x+13 000y=272 500
To simplify the second equation, divide both sides by 500.
23x+26y=545
Using the substitution method:
23x+26y=545
y=22 x
23x+26(22 x)=545
23x+572 26x=545
3x=545 572 = 27
x=9
He sells nine houses with an $11 500 commission.
7 It is easiest to let the third boy have mmarbles, in which case the sec-ond boy will have 2mmarbles and the first boy will have 2m 14.
(2m 14)+2m+m=71
5m 14=71
5m=85
m=17
The first boy has 20 marbles, the second boy has 34 and the third boy has 17 marbles, for a total of 71.
8 Let Belinda’s score beb.
Annie’s score will be 110% of Belinda’s or 1.1b.
Cassie’s will be 60% of their combined scores:
0.6(1.1b+b)= 0.6⇥2.1b = 1.26b
1.1b+b+1.26b= 504
3.36b= 504
b= 5.04
3.36
= 150
uncorrected
Belinda scores 150
Annie scores 1.1⇥150= 165
Cassie scores 0.6⇥(150+165)=189
9 Letrkm/h be the speed Kim can run. Her cycling speed will be (r+30) km/h. Her time cycling
will be 48+48÷3= 64 min.
Con-verting the times to hours (÷60) and using distance = speed ⇥ time
gives the following equation: r⇥ 48
60 +(r+30)⇥ 64 60 = 60 48r+64(r+30)= 60⇥60
48r+64r+1920= 3600
112r+1920= 3600
112r= 1680
r= 1680
112 = 15 She can run at 15 km/h
10 Letcg be the mass of a carbon atom and xg be the mass of an oxygen atom. (ois too confusing a symbol to use)
2c+6x=2.45⇥10 22
x= c
3 Use substitution.
2c+6⇥ c
3 =2.45⇥10 22 2c+2c=2.45⇥10 22
4c=2.45⇥10 22
c= 2.45⇥10
22 4
=6.125⇥10 23
x= c
3
= 6.125⇥10
23 3 ⇡2.04⇥10 23 The mass of an oxygen atom is 2.04⇥10 23g.
11 Let xbe the number of pearls. x
6 + x 3 +
x
5 +9= x 5x+10x+6x
30 +9= x
21x+270=30x
7x+90=10x
3x=90
x=30
There are 30 pearls.
12 Let the oldest receive $x.
The middle child receives $(x 12). The youngest child receives $ x 12
3 !
x+ x 12+ x 12
3 = 96 2x 12+ x 12
3 = 96 2x 12+ x
3 = 100 6x 36+x= 300
7x= 336
x 48
uncorrected
Oldest $48, Middle $35, Youngest $12
13 LetS be the sum of her marks on the first four tests.
Then S 4 = 88 ) S =352
Let xbe her mark on the fifth test. S +x
5 = 90 352+ x=450
x= 98
Her mark on the fifth test has to be 98%
14 LetN be the number of students in the class.
0.72N students have black hair After 5 leave the class there are 0.72N 5 students with black hair. There are nowN 5 students in the class.
Hence 0.72NN 5
5 =0.65
)0.72N 5=0.65(N 5)
) 0.72N =0.65N+1.75
) 0.07N =1.75
7N =175
N =25
There were originally 25 students
15 Amount of water in tank A at timet minutes=100 2t
Amount of water in tank B at timet minutes=120 3t
100 2t= 120 3t
t= 20
After 20 minutes the amount of water in the tanks will be the same.
16 Height candle A attminutes=10 5t
Height of candle B attminutes= 8 2t
a 10 5t=8 2t
3t=2
t= 2
3
)equal after 40 minutes. b 10 5t= 1
2(8 2t) 10 5t=4 t
4t=6
t= 3
2
)half the height after 90 minutes. c 10 5t=8 2t+1
10 5t=9 2t
3t=1
t= 1
3
)one centimetre more after 20 minutes.
17 Lettbe the time the motorist drove at 100 km/h
100t+90(10
3 t)= 320 100t+300 90t= 320
10t= 20
t= 2
Therefore the motorist travelled 200 km at 100 km/h
18 Let v km/h be Jarmila’s usual speed
uncorrected
Therefore distance travelled= 14v
3 km v+3 is the new speed and it takes 13
3 hours.
) 13
3 (v+3)= 14v
3 13(v+3)=14v
v=39
Her usual speed is 39 km/h
uncorrected
Solutions to Exercise 1E
1 Letkbe the number of kilometrestravelled in a day. The unlimited kilometre alternative will become more attractive when 0.32k+63>108. Solve for 0.32k+63= 108:
0.32k= 108 63 = 45
k= 45
0.32 =140.625
The unlimited kilometre alternative will become more attractive when you travel more than 140.625 km.
2 Letgbe the number of guests. Solve for the equality.
300+43g=450+40g
43g 40g=450 300
3g=150
g=50
Company A is cheaper when there are more than 50 guests.
3 Letabe the number of adults andcthe number of children.
45a+15c=525 000
a+c=15 000
Multiply the second equation by 15.
45a+15c=525 000 1
15a+15c=225 000 2
1 2:
30a=300 000
a=10 000
10 000 adults and 5000 children bought tickets.
4 Let $mbe the amount the contractor paid a man and $bthe amount he paid a boy.
8m+3b=2240
6m+18b=4200
Multiply the first equation by 6.
48m+18b=13 440 1
6m+18b=4200 2
1 2:
42m=9240
m=220
Substitute into the first equation: 8⇥220+3b= 2240
1760+3b= 2240
3b= 2240 1760 = 480
b= 160
He paid the men $220 each and the boys $160.
5 Let the numbers be xandy.
x+y=212 1
x y =42 2
1 + 2: 2x=254
x=127
127+y=212
y=85
The numbers are 127 and 85.
6 Let xL be the amount of 40% solu-tion andyL be the amount of 15% solution. Equate the actual substance.
uncorrected
0.4x+0.15y=0.24⇥700 =168
x+y=700
Multiply the second equation by 0.15.
0.4x+0.15y=168 1
0.15x+0.15y=105 2
1 2: 0.25x=63
x=63⇥4 =252
252+y=700
y=448
Use 252 L of 40% solution and 448 L of 15% solution.
7 Form two simultaneous equations.
x+y=220 1
x x
2 =y 40 x
2 y= 40 2
1 + 2: 3x
2 =180 x=120
120+y=220
y=100
They started with 120 and 100 marbles and ended with 60 each.
8 Let $xbe the amount initially invested at 10% and $ythe amount initially invested at 7%. This earns $31 000.
0.1x+0.07y=31 000
When the amounts are interchanged, she earns $1000 more, i.e. $32 000.
0.07x+0.1y=32 000
Multiply the first equation by 100 and the second equation by 70.
10x+7y= 3 100 000 1
4.9x+7y= 2 240 000 2
1 2:
5.1x= 860 000
x= 860 000
5.1 ⇡ 168 627.451 10⇥168 627.451+7y=3 100 000
1 686 274.51+7y=3 100 000
7y=1 413 725.49
y=201 960.78
The total amount invested is x+y=168 627.45+201 960.78
=$370 588.23 =$370 588
correct to the nearest dollar.
9 Letabe the number of adults and sthe number of students who attended.
30a+20s=37 000 1
a+ s=1600
20a+20s=1600⇥20
=32 000 2
1 2:
10a=5000
a=500
500+ s=1600
s=1100
500 adults and 1100 students attended the concert.
uncorrected
Solutions to Exercise 1F
1 a v= u+at= 15+2⇥5 = 25
b I = PrT
100
= 600⇥5.5⇥10
100
=330
c V =⇡r2h
=⇡⇥4.252⇥6 ⇡340.47 d S =2⇡r(r+h)
=2⇡⇥10.2⇥(10.2+15.6)
⇡1653.48 e V = 4
3⇡r2h
= 4⇡⇥3.58
2⇥11.4 3
⇡612.01 f s= ut+ 1
2at2
= 25.6⇥3.3+ 1
2 ⇥ 1.2⇥3.32 ⇡ 77.95
g T = 2⇡
s l g
= 2⇡⇥
r 1.45
9.8
= 2⇡⇥0.3846. . . ⇡ 2.42
h 1f = 1
v + 1 u
= 1
3 + 1 7 =
10 21 f = 21
10
=2.1
i c2 = a2+b2 = 8.82+3.42 = 89
c= p89
⇡ 9.43 j v2 = u2+2as
= 4.82+2⇥2.25⇥13.6 = 91.04
v= p91.04
⇡ 9.54
2 a v=u+at
v u =at
) a= v u
t
b S = n
2(a+l) 2S =n(a+l)
a+l= 2S
n ) l= 2S
n a
uncorrected
c A= 1
2bh 2A= bh
)b= 2A
h d P= I2R
P R = I2 ) I =±
r P R
e s=ut+ 1
2at2 s ut = 1
2at2 2(s ut)=at2
) a= 2(s ut)
t2
f E= 1
2mv2 2E= mv2
v2 = 2E m )v= ±
r 2E
m g Q= p2gh
Q2 = 2gh
)h= Q
2 2g
h xy z = xy+z
xy xy =z+z
2xy=2z
) x= 2z
2y
= z
y
i ax+c by = x b
ax+by=c(x b)
ax+by=cx bc
ax cx= bc by
x(a c)= b(c+y)
) x= b(c+y)
a c
= b(c+y)
c a j mxx b+b = c
mx+b= c(x b)
mx+b= cx bc
mx cx= bc b
x(m c)= b(c+1)
) x= b(c+1)
m c
3 a F = 9C
5 +32
= 9⇥28
5 +32
=82.4
b F = 9C
5 +32 F 32= 9C
5
9C =5(F 32)
) C = 5(F 32)
9 SubstituteF = 135.
C= 5(135 32)
9
= 515
9 ⇡ 57.22
uncorrected
4 a S =180(n 2) =180(8 2) =1080
b S=180(n 2)
S
180 =n 2 ) n= S
180 +2
= 1260
180 +2
=7+2=9
Polygon has 9 sides (a nonagon).
5 a V = 1
3⇡r2h
= 1
3 ⇥⇡⇥3.52⇥9 ⇡115.45 cm3
b V = 1
3⇡2h 3V = ⇡r2h
)h= 3V
⇡r2
= 3⇥210
⇡42 ⇡ 12.53 cm
c V = 1
3⇡r2h 3V = ⇡r2h
r2 = 3V ⇡h
) r=
r 3V ⇡h
=
r
3⇥262
⇡⇥10
⇡ 5.00 cm
6 a S = n
2(a+l)
= 7
2( 3+22)
=66.5
b S = n
2(a+l) 2S = n(a+l)
2S
n = a+l )a= 2S
n l
= 2⇥1040
13 156
= 4
c S = n
2(a+l) 2S = n(a+l)
)n= 2S
a+l = 2⇥110
25+ 5 = 11
There are 11 terms.
uncorrected
Solutions to Exercise 1G
1 a 2x 3 +
3x 2 =
4x+9x
6 = 13x 6 b 3a 2 a 4 = 6a a 4 = 5a 4 c 3h 4 + 5h 8 3h 2 =
6h+5h 12h
8 = h 8 d 3x 4 y 6 x 3 =
9x 2y 4x 12
= 5x 2y
12 e 3x + 2
y =
3y+2x
xy f x5
1 + 2 x =
5x+2(x 1)
x(x 1)
= 5x+2x 2
x(x 1)
= 7x 2
x(x 1) g x3
2 + 2 x+1 =
3(x+1)+2(x 2)
(x 2)(x+1) = 3x+3+2x 4
(x 2)(x+1) = 5x 1
(x 2)(x+1)
h x2x+3 x4x3 32
= 4x(x 3) 8x(x+3) 3(x+3)(x 3)
2(x+3)(x 3)
= 4x
2 12x 8x2 24x 3(x2 9)
2(x+3)(x 3)
= 4x
2 12x 8x2 24x 3x2+27
2(x+3)(x 3)
= 7x
2 36x+27
2(x+3)(x 3)
i x4
+1 +
3 (x+1)2 =
4(x+1)+3
(x+1)2 = 4x+4+3
(x+1)2 = 4x+7
(x+1)2
j aa2 + a
4 + 3a
8
= 8(a 2)+2a
2+3a2 8a
= 5a
2+8a 16 8a k 2x 6x2 4
5x =
10x2 (6x2 4) 5x
= 10x
2 6x2+4 5x
= 4x
2+4 5x
= 4(x
2+1) 5x
uncorrected
l x2
+4
3 x2+8x+16
= 2
x+4
3 (x+4)2 = 2(x+4) 3
(x+4)2 = 2x+8 3
(x+4)2 = 2x+5
(x+4)2
m x3 1 +
2 (x 1)(x+4) = 3(x+4)+2
(x 1)(x+4) = 3x+12+2
(x 1)(x+4) = 3x+14
(x 1)(x+4)
n x3 2
2 x+2 +
4 x2 4
= 3
x 2 2 x+2 +
4 (x 2)(x+2) = 3(x+2) 2(x 2)+4
(x 2)(x+2) = 3x+6 2x+4+4
(x 2)(x+2) = x+14
(x 2)(x+2)
o 5
x 2
3 x2+5x+6 +
2 x+3
= 5
x 2
3
(x+2)(x+3) +
2 x+3
= 5(x+3)(x+2) 3(x 2)+2(x 2)(x+2)
(x 2)(x+2)(x+3)
= 5(x
2+5x+6) 3x+6+2(x2 4)
(x 2)(x+2)(x+3)
= 5x
2+25x+30 3x+6+2x2 8
(x 2)(x+2)(x+3)
= 7x
2+22x+28
(x 2)(x+2)(x+3)
p x y 1
x y =
(x y)(x y) 1 x y
= (x y)
2 1 x y q x3
1
4x 1 x =
3 x 1 +
4x x 1
= 4x+3
x 1 r x3
2 + 2 2 x =
3 x 2
2x x 2
= 3 2x
x 2
2 a x2 2y ⇥
4y3 x =
4y3x2 2yx
=2xy2
b 3x2 4y ⇥
y2 6x =
3x2y2 24yx
= xy
8 c 4x3
3 ⇥ 12 8x4 =
48x3 24x4
= 2
x d x2
2y ÷ 3xy 6 = x2 2y ⇥ 6 3xy = 6x 2 6xy2 = x y2 e 4 x
3a ⇥ a2 4 x =
a2(4 x) 3a(4 x)
= a
3
uncorrected
f 2x+5 4x2+10x =
2x+5
2x(2x+5) = 1
2x g x2(x 1)2
+3x 4 =
(x 1)2 (x 1)(x+4) = x 1
x+4
h x2x x 6
3 =
(x 3)(x+2)
x 3
= x+2
i x2x2 5x+4 4x =
(x 1)(x 4) x(x 4)
= x 1
x j 5a2
12b2 ÷ 10a
6b = 5a2 12b2 ⇥
6b 10a = 30a 2b 120ab2 = a 4b k x x2 ÷ x22x24
= x 2
x ⇥ 2x2 x2 4
= x 2
x ⇥
2x2 (x 2)(x+2) = 2x
2 x(x+2) = 2x
x+2
l x+2 x(x 3) ÷
4x+8
x2 4x+3
= x+2
x(x 3)÷
4(x+2)
(x 1)(x 3)
= x+2
x(x 3)⇥
(x 1)(x 3) 4(x+2) = 1
x ⇥ x 1
4
= x 1
4x m
2x x 1 ÷
4x2 x2 1 =
2x x 1 ⇥
x2 1 4x2
= 2x
x 1 ⇥
(x 1)(x+1)
4x2
= 2x(x+1)
4x2
= x+1
2x n xx2 9
+2 ⇥
3x+6
x 3 ÷ 9 x
= (x 3)(x+3)
x+2 ⇥
3(x+2)
x 3 ⇥ x 9
= 3x(x 3)(x+3)(x+2)
9(x+2)(x 3) = x(x+3)
3
o 3x
9x 6 ÷ 6x2 x 2⇥
2 x+5 = 3x
3(3x 2)⇥ x 2
6x2 ⇥ 2 x+5 = 2x(x 2)
6x2(3x 2)(x+5)
= x 2
3x(3x 2)(x+5)
3 a x1 3 +
2 x 3 =
3 x 3
uncorrected
b x2 4 +
2 x 3 =
2(x 3)+2(x 4)
(x 4)(x 3)
= 2x 6+2x 8
x2 7x+12
= 4x 14
x2 7x+12 c x3
+4 +
2 x 3 =
3(x 3)+2(x+4)
(x+4)(x 3) = 3x 9+2x+8
x2+x 12
= 5x 1
x2+x 12 d
2x x 3 +
2 x+4 =
2x(x+4)+2(x 3)
(x 3)(x+4) = 2x
2+8x+2x 6 x2+x 12
= 2x
2+10x 6 x2+x 12
e 1
(x 5)2 + 2 x 5 =
1+2(x 5)
(x 5)2
= 1+2x 10
x2 10x+25
= 2x 9
x2 10x+25
f 3x
(x 4)2 + 2 x 4 =
3x+2(x 4)
(x 4)2
= 3x+2x 8
x2 8x+16
= 5x 8
x2 8x+16 g x1
3 2 x 3 =
1 x 3
= 1
3 x
h x2 3
5 x+4 =
2(x+4) 5(x 3)
(x 3)(x+4) = 2x+8 5x+15
x2+ x 12
= 23 3x
x2+x 12 i
2x x 3 +
3x x+3 =
2x(x+3)+3x(x 3)
(x 3)(x+3) = 2x
2+6x+3x2 9x x2 9
= 5x
2 3x x2 9
j 1
(x 5)2 2 x 5 =
1 2(x 5) (x 5)2
= 1 2x+10
x2 10x+25
= 11 2x
x2 10x+25
k 2x
(x 6)3
2 (x 6)2 =
2x 2(x 6) (x 6)3
= 2x 2x+12
(x 6)3
= 12
(x 6)3 l
2x+3
x 4
2x 4 x 3
= (2x+3)(x 3) (2x 4)(x 4)
(x 4)(x 3)
= (2x
2 3x 9) (2x2 12x+16) x2 7x+12
= 2x
2 3x 9 2x2+12x 16 x2 7x+12
= 9x 25
x2 7x+12
uncorrected
4 a p1 x+ p 2
1 x
=
p
1 xp1 x+2
p
1 x
= 1p x+2
1 x
= p3 x
1 x
b p 2 x 4 +
2 3 =
2px 4+6
3px 4 c p 3
x+4 +
2 p
x+4 =
5 p
x+4
d p 3 x+4 +
p x+4 = 3+
p
x+4px+4
p x+4 = 3p+ x+4
x+4 = px+7
x+4
e p3x3
x+4 3x
2px+4
= 3x
3 3x2px+4px+4 p
x+4 = 3x
3 3x2(x+4) p
x+4 = 3x
3 3x3 12x2 p
x+4 = 12x
2 p
x+4
f 3x3
2px+3 +3x
2px+3
= 3x
3+6x2px+3px+3 p
x+3 = 3x
3+6x2(x+3) p
x+3 = 3x
3+6x3+18x2 p
x+3 = 9x
3+18x2 p
x+3 = 9x
2(x+2) p
x+3
5 a (6x 3)13 (6x 3) 23
= (6x 3)13 1 (6x 3)23
= (6x 3)
1
3(6x 3)23 1 (6x 3)23
= 6x 3 1
(6x 3)23
= 6x 4
(6x 3)23
b (2x+3)13 2x(2x+3) 23
= (2x+3)13 2x (2x+3)23
= (2x+3)
1
3(2x+3)23 2x (2x+3)23
= 2x+3 2x
(2x+3)23
= 3
(2x+3)23
uncorrected
c (3 x)13 2x(3 x) 23
= (3 x)13 2x (3 x)23
= (3 x)
1
3(3 x)23 2x (3 x)23
= 3 x 2x
(3 x)23
= 3 3x
(3 x)23
Since (3 x)2 = (x 3)2, the answer is equivalent to 3 3x
(x 3)23 .
uncorrected
Solutions to Exercise 1H
1 a ax+n=max=m n
x= m n
a b ax+b=bx
ax bx = b
x(a b)= b
x= b
a b
This answer is correct, but to avoid a negative sign, multiply numerator and denominator by 1.
x= b
a b ⇥ 1 1
= b
b a c axb +c= 0
ax b = c ax= bc
x= bc
a d px=qx+5
px qx=5
x(p q)=5
x= 5
p q e mx+n=nx m
mx nx = m n
x(m n)= m n
x = m n
m n
= m+n
n m
f x1
+a =
b x
Take reciprocals of both sides: x+a= x
b x x
b = a x
b x= a x xb
b = a x xb
b ⇥b= ab x xb= ab
x(1 b)= ab
x= ab
1 b
g x ab = 2b
x+a
Take reciprocals of both sides: x a
b =
x+a
2b x a
b ⇥2b= x+a
2b ⇥2b 2(x a)= x+a
2x 2a= x+a
2x x =a+2a
x =3a
h x
m +n= x n +m x
m ⇥mn+n⇥mn= x
n ⇥mn+m⇥mn nx+mn2 =mx+m2n
nx mx=m2n mn2
x(n m)=mn(m n)
x= mn(m n)
n m
uncorrected
Note that n m = m+n = 1(m n)
) x= mn(n m)
n m
= mn
i b(ax+b)=a(bx a)
abx b2 =abx a2 abx abx= a2+b2
2abx= a2+b2
x= ( a
2+b2) 2ab
= a
2 b2 2ab
j p2(1 x) 2pqx= q2(1+x)
p2 p2x 2pqx= q2+q2x
p2x 2pqx q2x= q2 p2 x(p2+2pq+q2)= q2 p2 x= (q
2 p2) p2+2pq+q2
= p
2 q2 (p+q)2 = (p q)(p+q)
(p+q)2 = p q
p+q
k ax 1= x
b +2 x
a ⇥ab ab = x
b ⇥ab+2ab bx ab =ax+2ab
bx ax =2ab+ab
x(b a)=3ab
x= 3ab
b a
l
x a b+
2x a+b =
1
a2 b2
x(a b)(a+b)
a b +
2x(a+b)(a b) a+b =
(a+b)(a b) a2 b2
x(a+b)+2x(a b)=1 ax+bx+2ax 2bx=1
3ax bx=1 x(3a b)=1
x= 1
3a b
m p qx
t + p= qx t
p pt(p qx)
t + p⇥ pt=
pt(qx t) p p(p qx)+ p2t=t(qx t)
p2 pqx+ p2t=qtx t2
pqx qtx= t2 p2 p2t
qx(p+t)= (t2+ p2+ p2t)
x= t
2+p2+ p2t q(p+t) or
p2+ p2t+t2 q(p+t)
n x1
+a +
1 x+2a =
2 x+3a
Multiply each term by(x+a)(x+2a)(x+3a).
(x+2a)(x+3a)+(x+a)(x+3a)=2(x+a)(x+2a)
x2+5ax+6a2+x2+4ax+3a2=2x2+6ax+4a2
2x2+9ax+9a2=2x2+6ax+4a2
2x2 9ax 2x2 6ax=4a2 9a2
3ax= 5a2
x= 5a
2
3a
= 5a
3
uncorrected
2 ax+by= p;bx ay=q
Multiply the first equation byaand the second equation byb.
a2x+aby=ap 1
b2x aby =bp 2
1s+ 2:
x(a2+b2)=ap+bq
x= ap+bq
a2+b2 Substitute intoax+by= p:
a⇥ ap+bq
a2+b2 +by= p
a(ap+bq)+by(a2+b2)= p(a2+b2)
a2p+abq+by(a2+b2)= a2p+b2p by(a2+b2)= a2p+b2p a2p abq by(a2+b2)=b2p abq
y= b(bp aq)
b(a2+b2)
= bp aq
a2+b2 3 ax + y
b =1; x b +
y a = 1
First, multiply both equations byab, giving the following:
bx+ay=ab
ax+by=ab
Multiply the first equation byband the second equation bya:
b2x+aby=ab2 1
a2x+aby=a2b 2
1 2:
x(b2 a2)=ab2 a2b
x= ab
2 a2b b2 a2
= ab(b a)
(b a)(b+a) = ab
a+b
Substitute intobx+ay=ab:
b⇥ ab
a+b +ay= ab
ab2(a+b)
a+b +ay(a+b)= ab(a+b)
ab2+ay(a+b)= a2b+ab2 ay(a+b)= a2b+ab2 ab2
ay(a+b)= a2b
y= a
2b a(a+b) = ab
a+b
4 a Multiply the first equation byb.
abx+by= bc 1
x+by= d 2
1 2:
x(ab 1)= bc d
x= bc d
ab 1
= d bc
1 ab
It is easier to substitute in the first equation for x:
uncorrected
a⇥ abbc d1+y=c a(bc d)(ab 1)
ab 1 +y(ab 1)=c(ab 1)
abc ad+y(ab 1)=abc c y(ab 1)=abc c abc+ad y(ab 1)= c+ad
y= ad c ab 1
= c ad
1 ab b Multiply the first equation byaand
the second equation byb.
a2x aby =a3 1
b2x aby =b3 2
1 2:
x(a2 b2)=a3 b3 x= a
3 b3 a2 b2
= (a b)(a
2+ab+b2) (a b)(a+b) = a
2+ab+b2 a+b
In this case it is easier to start again, but eliminate x.
Multiply the first equation byband the second equation bya.
abx b2y= a2b 3
abx a2y= ab2 4
3 4:
y( b2+a2)= a2b ab2
y(a2 b2)= ab(a b)
y= ab(a b)
a2 b2
= ab(a b)
(a b)(a+b) = ab
a+b
c Add the starting equations: ax+by+ax by= t+s
2ax= t+s
x= t+s
2a Subtract the starting equations:
ax+by (ax by)= t s
2by= t s
y= t s
2b
d Multiply the first equation byaand the second equation byb.
a2x+aby=a3+2a2b ab2 1 b2x+aby=a2b+b3 2
1 2:
x(a2 b2)=a3+a2b ab2 b3 x= a
3+a2b ab2 b3 a2 b2
= a
2(a+b) b2(a+b) a2 b2
= (a
2 b2)(a+b) a2 b2
=a+b
Substitute into the second, simpler equation.
b(a+b)+ay=a2+b2
ab+b2 +ay=a2+b2
ay=a2+b2 ab b2
ay=a2 ab
y= a
2 ab a
=a b
e Rewrite the second equation, then multiply the first equation byb+c
uncorrected
and the second equation byc. (a+b)(b+c)x+c(c+c)y
= bc(b+c) 1
acx+c(b+c)y
= abc 2
1 2:
x((a+b)(b+c) ac) = bc(b+c)+abc
x(ab+ac+b2+bc ac) = bc(b+c+a)
x(ab+b2+bc)=bc(a+b+c)
xb(a+b+c)=bc(a+b+c)
x= bc(a+b+c)
b(a+b+c) =c
Substitute into the first equation. (It has the simpleryterm.)
c(a+b)+cy=bc
ac+bc+cy=bc
cy=bc ac bc
cy= ac
y= ac
c
= a
f First simplify the equations. 3x 3a 2y 2a= 5 4a
3x 2y= 5 4a +3a+2a
3x 2y= a+5 1
2x+2a+3y 3a= 4a 1
2x+3y= 4a 1
2a+3a
2x+3y= 5a 1 2
Multiply1 by 3 and 2 by 2.
9x 6y=3a+15 3
4x+6y=10a 2 4
3 + 4:
13x=13a+13
x=a+1
Substitute into 2: 2(a+1)+3y=5a 1
2a+2+3y=5a 1
3y=5a 1 2a 2
3y=3a 3
y=a 1
5 a s=ah =a(2a+1)
b Makehthe subject of the second equation.
h=a(2+h) =2a+ah
h ah =2a
h(1 a)=2a
h= 2a
1 a
Substitute into the first equation. s=ah
=a⇥ 2a
1 a
= 2a
2
1 a
uncorrected
c h+ah=1
h(1+a)=1
h= 1
(1+a) =
1 a+1
as=a+h =a+ 1
a+1 = a(a+1)+1
a+1 = a
2+a+1 a+1
s= a
2+a+1 a(a+1)
d Makehthe subject of the second equation.
ah=a+h
ah h =a
h(a 1)=a
h= 1
a 1
Substitute into the first equation. as= s+h
as= s+ a
a 1 as s= a
a 1 s(a 1)= a
a 1 s(a 1)(a 1)= a(a 1)
a 1 s(a 1)2 =a
s= a
(a 1)2
e s=h2+ah
=(3a2)2 +a(3a2) =9a4+3a3 =3a3(3a+1)
f as= a+2h = a+2(a s) = a+2a 2s
as+2s= 3a
s(a+2)= 3a
s= 3a
a+2
g s=2+ah+h2 =2+a
✓ a 1
a ◆
+
✓
a 1
a ◆2
=2+a2 1+a2 2+ 1
a2
=2a2 1+ 1
a2
h Makehthe subject of the second equation.
as+2h= 3a
2h= 3a as
h= 3a as
2
Substitute into the first equation. 3s ah= a2
3s a(3a as)
2 = a2
6s a(3a as)= 2a2
6s 3a2+a2s= 2a2
a2s+6s= 2a2+3a2 s(a2+6)= 5a2
s= 5a
2 a2+6
uncorrected
Solutions to Exercise 1I
Use your CAS calculator to find the solu-tions to these problems. The exact method will vary depending on the calculator used.
1 a x= a b
b x= 7
c x= a±
p
a2+4ab 4b2 2
d x= a+c
2
2 a (x 1)(x+1)(y 1)(y+1)
b (x 1)(x+1)(x+2)
c (a2 12b)(a2+4b) d (a c)(a 2b+c)
3 a axy+b=(a+c)y
bxy+a=(b+c)y
Dividing byyyields: ax+ b
y = a+c bx+ a
y = b+c letn= 1
y and the equations become: ax+bn= a+c
bx+an= b+c
) x= a+b+c
a+b
y= a+b
c
b x(b c)+by c =0
y(c a) ax+c=0
(b c)x+by=c
ax+(c a)y= c
) x= (a b c)
a+b c
y= a b+c
a+b c
uncorrected
Solutions to technology-free questions
1 a (x3)4 = x3⇥4= x12
b (y 12)34 = y 12⇥34
= y 9
c 3x32 ⇥ 5x4 =(3⇥ 5)x32+4
= 15x112 d (x3)43 ⇥x 5 = x3⇥43 ⇥x 5
= x4 5 = x 1
2 32⇥1011⇥12⇥10 5
= (32⇥12)⇥1011 5 = 384⇥106
= 3.84⇥108
3 a 3x 5 +
y 10
2x 5 =
6x+y 4x
10
= 2x+y
10 b 4x 7y = 4y 7x
xy c x5
+2 +
2 x 1 =
5(x 1)+2(x+2)
(x+2)(x 1) = 5x 5+2x+4
(x+2)(x 1) = 7x 1
(x+2)(x 1)
d x3
+2 +
4 x+4 =
3(x+4)+4(x+2)
(x+2)(x+4) = 3x+12+4x+8
(x+2)(x+4) = 7x+20
(x+2)(x+4)
e x5x+4 + 4x
x 2
5 2
= 10x(x 2)+8x(x+4) 5(x+4)(x 2)
2(x+4)(x 2)
= 10x
2 20x+8x2+32x 5(x2+2x 8)
2(x+4)(x 2)
= 10x
2 20x+8x2+32x 5x2 10x+40
2(x+4)(x 2)
= 37x
2+2x+40
2(x+4)(x 2)
f x3 2
6 (x 2)2 =
3(x 2) 6 (x 2)2
= 3x 6 6
x 2
= 3x 12
x 2
= 3(x 4)
x 2
4 a x+5 2x 6 ÷
x2+5x 4x 12
= x+5
2x 6 ⇥
4x 12 x2+5x
= x+5
2(x 3)⇥
4(x 3) x(x+5) = 4
2x = 2 x
uncorrected
b x3x
+4 ÷
12x2 x2 16
= 3x
x+4 ⇥
x2 16 12x2
= 3x
x+4 ⇥
(x 4)(x+4)
12x2
= 3x(x 4)
12x2
= x 4
4x c xx2 4
3 ⇥
3x 9 x+2 ÷
9 x+2 = x
2 4 x 3 ⇥
3x 9 x+2 ⇥
x+2
9
= (x 2)(x+2)
x 3 ⇥
3(x 3) x+2
⇥ x+92
= (x+2)(x 2)
3 =
x2 4 3 d 4x+20
9x 6 ⇥ 6x2 x+5 ÷
2 3x 2
= 4(x+5)
3(3x 2) ⇥ 6x2 x+5 ⇥
3x 2 2
= 4⇥6x
2 3⇥2 =4x2
5 Lettseconds be the required time. The number of red blood cells to be replaced is 1
2 ⇥5⇥1012= 2.5⇥1012 2.5⇥106⇥t= 2.5⇥1012
t= 2.5⇥10
12 2.5⇥106
= 106
Time= 106seconds
= 106÷3600÷24 days
⇡ 11.57 or 11 31 54 days
6 1.5⇥108
3⇥106 =0.5⇥102
=50 times further
7 Letgbe the number of games the team lost. They won 2ggames and drew one third of 54 games, i.e. 18 games.
g+2g+18= 54
3g= 54 18 = 36
g= 12
They have lost 12 games.
8 Letbbe the number of blues CDs sold. The store sold l.lb clas-sical and 1.5(b+1.1b) heavy
metal CDs, totalling 420 CDs. b+1.1b+1.5⇥2.1b= 420
5.25b= 420
b= 420
5.25
= 80
1.1b= 1.1⇥80=88
1.5⇥2.1b= 1.5⇥2.1⇥80 = 252
80 blues, 88 classical and 252 heavy metal (totalling 420)
9 a V = ⇡r2h = ⇡⇥52⇥12
= 300⇡⇡ 942 cm3
uncorrected
b h= V
⇡r2
= 585
⇡⇥52
= 117
5⇡ ⇡ 7.4 cm c r2 = V
⇡h
r =
r V
⇡h (use positive root)
=
r 786
⇡⇥6
=
r 128
⇡ ⇡40.7 cm 10 a xy+ax=b
x(y+a)=b
x = b
a+y
b ax + b
x =c ax
x + bx
x =cx a+b=cx
x= a+b
c
c ax = x
b +2 xab
a = xab
b +2ab bx=ax+2ab
bx ax =2ab
x(b a)=2ab
x= 2ab
b a
d
a dx d +b=
ax+d
b bd(a dx)
d +bd⇥b=
bd(ax+d)
b b(a dx)+b2d= d(ax+d)
ab bdx+b2d= adx+d2
bdx adx= d2 ab b2d
x(bd+ad)= (ab+b2d d2)
x= (ab+b
2d d2) (bd+ad) = ab+b
2d d2 bd+ad
11 a pp
+q+
q p q =
p(p q)+q(p+q)
(p+q)(p q) = p
2 qp+qp+q2 p2 pq+ pq q2
= p
2+q2 p2 q2 b 1x xy y2y 2 = (xy y
2) 2xy x(xy y2)
= xy y
2 x2y xy2
= y( x y)
xy(x y)
= x y
x(x y)
= x+y
x(y x) c x2x+x 6
+1 ⇥
2x2+x 1 x+3 = (x 2)(x+3)
x+1 ⇥
(x+1)(2x 1)
x+3 = (x 2)(2x 1)
uncorrected
d 2a 2a+b ⇥
2ab+b2
ba2
= 2a
2a+b ⇥
b(2a+b)
ba2
= 2ab
ba2
= 2
a
12 LetA’s age bea,B’s age bebandC’s age bec.
a=3b
b+3=3(c+3)
a+15=3(c+15)
Substitute foraand simplify: b+3=3(c+3)
b+3=3c+9
b=3c+6 1
3b+15=3(c+15)
3b+15=3c+45
3b=3c+30
b=c+10 2
1 = 2:
3c+6=c+10
3c c =10 6
2c=4
c=2
b=3⇥2+6 =12
a=3⇥12 =36
A,BandC are 36, 12 and 2 years old respectively.
13 a Simplify the first equation: a 5= 1
7(b+3) 7(a 5)=b+3
7a 35=b+3
7a b =38
Simplify the second equation: b 12= 1
5(4a 2) 5(b 12)=4a 2
5b 60=4a 2
4a+5b=58
Multiply the first equation by 5, and add the second equation.
35a 5b=190 1
4a+5b=58 2
1 + 2: 31a=248
a=8
Substitute into the first equation: 7⇥8 b=38
56 b=38
b=56 38 =18
b Multiply the first equation by p. (p q)x+(p+q)y=(p+q2)
p(p q)x+ p(p+q)y= p(p+q2)
1 Multiply the second by (p+q).
qx py=q2 pq
q(p+q)x p(p+q)y =(p+q)(q2 pq) 2 1 + 2:
(p(p q)+q(p+q))x
= p(p+q)2+(p+q)(q2 pq)
uncorrected
(p2 pq+ pq+q2)x = p(p2+2pq+q2)
+ pq2 p2q+q3 pq2
(p2 +q2)x
= p3+2p2q+ pq2 p2q+q3 = p3+ p2q+ pq2+q3
= p2(p+q)+q2(p+q) = (p+q)(p2+q2)
x= p+q
Substitute into the second equa-tion, factorising the right side.
q(p+q) py=q2 pq
pq+q2 py=q2 pq
py=q2 pq pq q2
py= 2pq
y= 2pq
p
=2q
14 Time = distance
speed
Remainder= 50 7 7=36 km
7 x +
7 4x +
36 6x+3 = 4
7 x +
7 4x +
12 2x+1 = 4
(4x(2x+1))⇥ 7
x + 7 4x +
12 2x+1
!
= 4⇥4x(2x+1)
28(2x+1)+7(2x+1)+48x = 16x(2x+1)
56x+28+14x+7+48x = 32x2+16x
56x+28+14x+7+48x
32x2 16x= 0
32x2+102x+35=0
32x2 102x 35=0 (2x 7)(16x+5)=0
2x 7=0 or 16x+5= 0
x>0,so 2x 7= 0
x=3.5
15 a 2n2⇥6nk2÷3n= 2n2⇥6nk2 3n
= 12n
3k2 3n
= 4n2k2
b 8c2x3y 6a2b3c3 ÷
1 2xy 15abc2
= 8c
2x3y 6a2b3c3 ÷
xy 30abc2
= 8c
2x3y 6a2b3c3 ⇥
30abc2 xy
= 240abc
4x3y 6a2b3c3xy
= 40cx
2 ab2
16 x+5
15
x 5 10 =1+
2x 15 30(x+5)
15
30(x 5)
10 =30⇥ ✓
1+ 2x
15 ◆
2(x+5) 3(x 5)=30+4x
2x+10 3x+15=30+4x
2x 3x 4x=30 10 15
5x=5
x= 1
uncorrected
Solutions to multiple-choice questions
1 A 5x+2y= 02y= 5x
y x =
5 2
2 A Multiply both sides of the second equation by 2.
3x+2y=36 1
6x 2y=24 2
1 + 2:
9x=60
x= 20
3 3⇥ 203 y=12
20 y=12
y=8
3 C t 9= 3t 17
t 3t= 9 17
2t= 8
t= 4
4 A m= n p
n+ p
m(n+ p)=n p
mn+mp=n p
mp+ p=n mn
p(m+1)=n(1 m)
p= n(1 m)
1+m
5 B 3 x 3
2 x+3 =
3(x+3) 2(x 3)
(x 3)(x+3) = 3x+9 2x+6
x2 9
= x+15
x2 9
6 E 9x2y3÷15(xy)3 = 9x2y3 15(xy)3
= 9x
2y3 15x3y3
= 9
15x
= 3
5x 7 B V = 1
3h(l+w) 3V = h(l+w)
3V = hl+hw
hl= 3V hw
l= 3V hw
h
= 3V
h w
8 B (3x2y3)2 2x2y =
9x4y6 2x2y
= 9x
2y5 2
= 9
2x2y5
uncorrected
9 B Y = 80%⇥Z = 4
5Z X= 150%⇥Y = 3
2Y
= 3
2⇥ 4Z
5
= 12Z
10
= 1.2Z
= 20% greater thanZ
10 B Let the other number ben. x+n
2 = 5x+4 x+n= 2(5x+4)
= 10x+8
n= 10x+8 x = 9x+8
uncorrected
Solutions to extended-response questions
1 Jack cycles 10xkm.Benny drives 40xkm.
a Distance= speed⇥time
) time= distance
speed ) time taken by Jack= 10x
8
= 5x
4 hours b Time taken by Benny = 40x
70
= 4x
7 hours c Jack’s time–Benny’s time= 5x
4 4x
7
= (35 16)x
7
= 19x
28 hours d i If the di↵erence is 30 mins= 1
2 hour then 19x
28 = 1 2 ) x= 14
19
= 0.737 (correct to three decimal places)
ii Distance for Jack=10⇥ 14
19
= 140
19
=7 km (correct to the nearest km)
Distance for Benny=40⇥ 14
19
= 560
19
=29 km (correct to the nearest km)
uncorrected
2 a Dinghy is filling with water at a rate of 27 000 9 000 =18 000 cm3per minute.
b Aftertminutes there are 18 000tcm3water in the dinghy, i.e. V = 18 000t
c V =⇡r2his the formula for the volume of a cylinder
) h= V
⇡r2
= 18 000t
⇡r2
The radius of this cylinder is 40 cm ) h= 18 000t
1600⇡ = 45t
4⇡
i.e. the heighthcm water at timetis given byh= 45t
4⇡ d Whent=9, h= 45⇥9
4⇡ ⇡ 32.228. . .
The dinghy has filled with water, beforet= 9, i.e. Sam is rescued after the dinghy
completely filled with water.
3 a Let Thomas haveacards. Therefore Henry has 5a
6 cards, George has 3a
2 cards, Sally has (a 18) cards and Zeb has a
3 cards. b 3a
2 +a 18+ a 3 =a+
5a 6 +6
c) 9a+6a 108+2a=6a+5a+36
) 6a=144
) a=24
Thomas has 24 cards, Henry has 20 cards, George has 36 cards, Sally has 6 cards and Zeb has 8 cards.
4 a F = 6.67⇥10
11⇥200⇥200 122
=1.852. . .⇥10 8
=1.9⇥10 8 (correct to two significant figures)
uncorrected
b m1 = Fr 2
m2⇥6.67⇥10 11
= Fr
2⇥1011 6.67m2 c IfF =2.4⇥104
r =6.4⇥106 andm2 =1500
m1 = 2.4⇥10
4⇥(6.4⇥106)2⇥1011 6.67⇥1500
=9.8254. . .⇥1024
The mass of the earth is 9.8⇥1024kg (correct to two significant figures). 5 a V =3⇥103⇥6⇥103⇥d
=18⇥106d
b Whend= 30, V =18⇥106⇥30 = 540 000 000
= 5.4⇥108
The volume of the reservoir is 5.4⇥108 m3.
c E= kVh
1.06⇥1015= k⇥200⇥5.4⇥108 k= 1.06⇥10
15 200⇥5.4⇥108
= 9.81. . .⇥103
k= 9.81⇥103correct to three significant figures.
d E= (9.81⇥103)⇥5.4⇥108⇥250
= 1.325⇥1015correct to four significant figures.
The amount of energy produced is 1.325⇥1015J. e Lettbe the time in seconds.
5.2⇥t =5.4⇥108
t =103.846 153 8
) number of days=103.846 153 8÷(24⇥60⇥60) =1201.92. . .
The station could operate for approximately 1202 days.
uncorrected
CAS calculator techniques for Question 5
5 b Calculations involving scientific notation andsignificant figures can be accomplished with the aid of a graphics calculator.
Whend= 30, V =18⇥106⇥30 =540 000 000
This calculation can be completed as shown here.
T1: Press c!5:Settings!2:Document Settingsand change the Exponential Format to Scientific. Click on Make Default.
Return to the Calculator application. Type18⇥10^6⇥30or18i6⇥30
CP: In the Main application tap !Basic Format
Change the Number Format to Sci2 Type18⇥10^6⇥30
c T1: Press c!5: Settings!2: Document Settingsand change the Display Digits to Float 3. Click on Make Default.
Return to the home screen and press and complete as shown. CP: tap !Basic Format
Change the Number Format to Sci3 Complete calculation as shown d The calculation is as shown. T1: Display Digits
is Float 4CP: Number Format is Sci4 Simply type⇥5.4⇥10^8⇥25
uncorrected
6 LetR1cm andR2 cm be the radii of the inner circles. ) Yellow area =⇡R21
Blue area =⇡R22 ⇡R21
Red area =100⇡ ⇡R22
) 100⇡ ⇡R22 =⇡R22 ⇡R21 = ⇡R21
Firstly, ⇡R22 ⇡R21 =⇡R21
implies R2
2 =2R21 1
and 100⇡ ⇡R2
2 =⇡R22 ⇡R21
implies 100=2R22 R21 2
Substitute from 1in 2
) 100= 4R21 R21
100= 3R21
and R1 = 10p
3
= 10
p 3 3
⇣
Note : R2
2 = 2003 ⌘
The radius of the innermost circle is 10 p
3
3 cm.
R1 R
2
10 cm
Blue Yellow
Red
7 IfC = F,
F = 5
9(F 32) 9F =5F 160
)4F = 160
) F = 40
Therefore 40 F = 40 C.
uncorrected
8 Let xkm be the length of the slope. ) time to go up= x
15 ) time to go down= x
40 ) total time= x
15 + x 40
= 11x
120 ) average speed= total distance
total time
= 2x÷ 11x
120
= 2x⇥ 120
11x
= 240
11
⇡ 21.82 km/h
xkm
9 1 litre =1000 cm3
a Volume=Volume of cylinder+Volume of hemisphere =⇡r2h+ 2
3⇡r3 It is known thatr+h=20
) h=20 r
b i Volume=⇡r2(20 r)+ 2
3⇡r3
=20⇡r2 ⇡r3+ 2
3⇡r3
=20⇡r2 ⇡
3r3 ii If Volume= 2000 cm3
then 20⇡r2 ⇡
3r3 = 2000
Use a CAS calculator to solve this equation forr, given that 0<r <20. This givesr= 5.943999· · ·
Thereforeh= 20 r
= 20 5.943 99. . .
= 14.056 001. . .
uncorrected
The volume is two litres whenr =5.94 andh=14.06, correct to two decimal
places.
10 a Let xandybe the amount of liquid (in cm3) taken from bottlesAandBrespectively. Since the third bottle has a capacity of 1000 cm3,
x+y=1000 1
Now x= 2
3xwine+ 1
3xwater
and y= 1
6ywine+ 5
6ywater
) 2
3x+ 1 6y=
1 3x+
5
6ysince the proportion of wine and water must be the same. ) 4x+y=2x+5y
) 2x=4y
) x=2y
From 2 2y+y=1000
) y= 1000
3 andx = 2000
3 Therefore, 2000
3 cm3 and 1000
3 cm3 must be taken from bottlesAandBrespectively so that the third bottle will have equal amounts of wine and water, i.e. 2
3Lfrom A and 1
3Lfrom B
b x+y=1000 1
1 3x+
3 4y=
2 3x+
1 4y
) 4x+9y=8x+3y
) 6y=4x
) x= 3
2y 2
From 1 3
2y+y=1000
) y= 2
5 ⇥1000
=400
) x=600
uncorrected
Therefore, 600 cm3and 400 cm3 must be taken from bottlesAandBrespectively so that the third bottle will have equal amounts of wine and water, i.e. 600 mL from A and 400 mL from B
c
x+y=1000 1
m m+nx+
p p+qy=
n m+nx+
q p+qy
) m(p+q)x+ p(m+n)y=n(p+q)x+q(m+n)y
) (m(p +q) n(p+q))x=(q(m+n) p(m+n))y
) (m n)(p+q)x=(q p)(m+n)y
) x= (m+n)(q p)
(m n)(p+q)y, m,n,p, q 2
From 1 (m+n)(q p)
(m n)(p+q)y+y=1000
) (m+n)(q p)+(m n)(p+q)
(m n)(p+q) y=1000
) mq mp+nq np+mp+mq np nq
(m n)(p+q) y=1000
) 2(mq np)
(m n)(p+q)y=1000
) y= 500(m n)(p+q)
mq np ,
mq,np From 1 x= (m+n)(q p)
(m n)(p+q) ⇥
500(m n)(p+q)
mq np
= 500(m+n)(q p)
mq np ,
n q ,
q p Therefore, 500(m+n)(q p)
mq np cm3 and
500(m n)(p+q)
mq np cm3must be taken from bottlesAandBrespectively so that the third bottle will have equal amounts of wine and water. In litres this is (m+n)(q p)
2(mq np) litres from A and
(m n)(p+q)
2(mq np) litres from B. Also note that n
m 1 and q
p 1 or n
m 1 and q p 1.
uncorrected
11 a 20 h 20 =
r 10 ) 10(20 h)=20r
) 200 10h=20r
) 20 h=2r
) h=20 2r =2(10 r)
b V =⇡r2h
=2⇡r2(10 r)
c Use CAS calculator to solve the equation 2⇡x2(10 r)=500, given that 0< r< 10. This givesr =3.49857. . .orr =9.02244. . .
Whenr= 3.498 57. . . , h=2(10 3.498 57. . .)
= 13.002 85. . .
Whenr= 9.022 44. . . , h=2(10 9.022 44. . .) = 1.955 11. . .
Therefore the volume of the cylinder is 500 cm3 whenr=3.50 andh=13.00 or whenr= 9.02 andh=1.96, correct to two decimal places.
uncorrected
Chapter 2 – Number systems and sets
Solutions to Exercise 2A
1
3 1 x
5
4 B
A 4
3 1
5 2
a A0 ={4}
b B0 ={1,3,5}
c A∪B={1,2,3,4,5}, orξ
d (A∪B)0 =∅
e A0∩B0 = ∅
2
3
15 9
1 5 7 11 13
4 2 6 12
14 8 10
16
P Q
x
a P0 ={1,2,4,5,7,8,10,11,13,14,16}
b Q0 ={1,3,5,7,9,11,13,15}
c P∪Q=
{2,3,4,6,8,9,10,12,14,15,16}
d (P∪Q)0 ={1,5,7,11,13}
e P0∩Q0 = {1,5,7,11,13}
3
A
B
1 3 7
5 9 11 8 12
4 6 2
10
ξ
a A0 ={1,2,3,5,6,7,9,10,11}
b B0 ={1,3,5,7,9,11}
c A∪B={2,4,6,8,10,12}
d (A∪B)0 ={1,3,5,7,9,11}
e A0∩B0 ={1,3,5,7,9,11}
4 11 13 14 17 18 19 21 22 23
12 16
24
P
x
Q
20 10
25 15
a P0 =
{10,11,13,14,15,17,18,19,21,22,23,25}
b Q0 =
{11,12,13,14,16,17,18,19,21,22,23,24}
c P∪Q={10,12,15,16,20,24,25}
d (P∪Q)0 =
{11,13,14,17,18,19,21,22,23}
e P0∩Q0 =
{11,13,14,17,18,19,21,22,23}
5
r w X
p x
Y
t s
q
v u
a X0 = {p, q, u, v}
b Y0 ={p, r, w}
c X0∩Y0 ={p}
d X0∪Y0 ={p, q, r, u, v, w}
e X∪Y = {q, r, s, t, u, v, w}
uncorrected
f (X∪Y)0 ={p}candfare equal.
6 5 7 9 11
1 3
2 6 124
X
x
Y
8 10
a X0 ={5,7,8,9,10,11}
b Y0 ={1,3,5,7,9,11}
c X0 ∪Y0 ={1,3,5,7,8,9,10,11}
d X0 ∩Y0 ={5,7,9,11}
e X∪Y ={1,2,3,4,6,8,10,12}
f (X∪Y)0 ={5,7,9,11}dandfare equal.
7 a ξ A B
A¢
b ξ A B
B¢
c
B
A
ξ
A¢ Ç B¢
d
B
A
ξ
A¢È B¢
e ξ
AÈ B A B
f ξ
(AÈ B)¢
B
A
8
G
A
B L
A NE R x
a A0 ={R}
b B0 ={G, R}
c A∩B={L, E, A, N}
d A∪B={A, N, G, E, L}
e (A∪B)0 ={R}
f A0∪B0 ={G, R}
9
B
A I
C A T
S E H
M
ξ
a A0 ={E, H, M, S} b B0 ={C, H, I, M}
c A∩B={A, T} d (A∪B)0 ={H, M}
e A0∪B0 ={C, E, H, I, M, S} f A0∩B0 ={H, M}
uncorrected
Solutions to Exercise 2B
1 a Ye
b Yes
c Yes
2 a The sum may be rational or irrational, for instance, √2+ √3 is irrational; √
2+(3− √2)= 3 is rational.
b The product may be rational or irra-tional. For instance, √2× √3= √6 is irrational; √2×3√2=6 is rational.
c The quotient may be rational or irrational. For instance
√ 2 √
3 is irrational; 3
√ 2 √
2 =3 is rational.
3 a 0.45= 45 100 =
9 20
b 0.˙2˙7= 0.272727. . . 0.˙2˙7×100= 27.272727. . .
0.˙2˙7×99= 27
∴ 0.˙2˙7= 27 99 =
3 11
c 0.12= 12 100 =
3 25 d
0.˙28571 ˙4= 0.285714285714. . . 0.˙28571 ˙4×106 = 285714.285714. . . 0.˙28571 ˙4×(106−1)= 285714
∴ 0.˙28571 ˙4= 285714 999999 =
2 7
e 0.˙3˙6=0.363636. . . 0.˙3˙6×100=36.3636. . .
0.˙3˙6×99=36
∴ 0.˙3˙6= 36 99 =
4 11 f 0.˙2=0.22222. . . 0.˙2×10=2.2222. . .
0.˙2×9=2
∴ 0.˙2= 2 9
4 a 2 7 =7
2.000000. . .
=0.2857142857. . .
=0.˙28571 ˙4
b 5
11 =11
5.000000. . .
=0.454545. . .
=0.˙4˙5
c 7
20 =20
7.00
=0.35
d 4
13 =13
4.000000. . .
=0.30769230. . .
=0.˙30769˙2
e 1
17 =17
1.00000000000000000. . .
=0.0588235294117647058. . .
=0.˙058823529411764˙7
uncorrected
5 a
−2 −1 0 1 2 3 4 5
b
−3 −2 −1 0 1 2 3 4
c
−3 −2 −1 0 1 2 3 4
d
−2 −1 0 1 2 3 4 5
e
−2 −1 0 1 2 3 4 5
6 a (−∞,3)
b [−3,∞)
c (−∞,−3]
d (5,∞)
e [−2,3)
f [−2,3]
g (−2,3]
h (−5,3)
uncorrected
Solutions to Exercise 2C
1 a 8
b 8
c 2
d −2
e −2
f 4
2 a |x−1|= 2 Case 1: If x≥1
x−1= 2
x= 3 Case 2: If x<1
1−x= 2
x= −1
b |2x−3|=4
Case 1: If x≥ 3 2 2x−3=4
x= 7
2 Case 2: If x< 3
2 3−2x=4
x=−1 2
c |5x−3|=9
Case 1: If x≥ 3 5 5x−3=9
x= 12
5 Case 2: If x< 3
5 3−5x=9
x=−6 5 d |x−3|=9
Case 1: If x≥3
x−3=9
x=12 Case 2: If x<3
3−x=9
x=−6
e |x−3|=4 Case 1: If x≥3
x−3=4
x=7 Case 2: If x<3
3−x=4
x=−1
unc
